A thick cylinder with 10 mm internal diameter and 20 mm external diameter, is subjected to an internal fluid pressure of 60 MPa. The hoop stress at the inner surface is

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UPPSC AE Mechanical 2022 Official Paper I (Held on 29 May 2022)
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  1. 140 MPa
  2. -60 MPa
  3. 100 MPa
  4. 40 MPa

Answer (Detailed Solution Below)

Option 3 : 100 MPa
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Concept:-

When a thick-walled cylinder is subjected to internal pressure hoop stress, radial stress and longitudinal stress are developed.

The stress in the circumferential direction or hoop stress or tangential stress at a point in a tube or cylinder can be expressed as, 

\(σ_c= \left[\frac{(P_i r_i^2-P_o r_o^2)}{(r_o^2- r_i^2)}\right] -\left[ \frac{r_i^2 r_o^2(P_o-P_i)}{r^2(r_o^2- r_i^2)}\right]\;\)

where Pi  = Internal pressure in the cylinder (MPa), P= External pressure in the cylinder (MPa), σ= Circumferential or Hoop stress (MPa)

r= Internal radius of cylinder (mm), r= Outer radius of cylinder (mm)

r = Radius to the point at which hoop stress is to be calculated (mm)

F1 M.J Madhu 04.04.20 D4

Calculation:

Given:

P= 60 MPa, P= 0, r= 5 mm, r= 10 mm.

We have to calculate hoop stress at the inner surface so, r = r= 5 mm

The formula reduces to

\(\sigma _c= P_i\left[\frac{r_i^2 + r_o^2}{r_o^2-r_i^2}\right]\)

\(\sigma_c= 60\left[\frac{10^2+ 5^2}{10^2- 5^2}\right] = 100 MPa\)

Note: Hoop stress is maximum on the inner side of the cylinder and minimum on the outer side and radial stress is maximum and equal to internal pressure on the inner side and zeroes on the outer side of the cylinder.

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